# What are the critical values, if any, of f(x)= (x^2+6x-7)^2?

Nov 19, 2015

The critical values for $f$ are $- 7 , - 3 , \text{and } 1$

#### Explanation:

Use the power rule and the chain rule to find:

$f ' \left(x\right) = 2 \left({x}^{2} + 6 x - 7\right) \left(2 x + 6\right)$.

Since, $f \left(x\right)$ and $f ' \left(x\right)$ are polynomials, the critical values for $f$ are the zeros of $f '$

Solve $f ' \left(x\right) = 2 \left({x}^{2} + 6 x - 7\right) \left(2 x + 6\right) = 0$ by solving

${x}^{2} + 6 x - 7 = 0$ or $2 x + 6 = 0$. So,

$\left(x + 7\right) \left(x - 1\right) = 0$ or $x = - 3$

$x = - 7 , 1 , \text{or} - 3$

The critical values for $f$ are $- 7 , - 3 , \text{and } 1$